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Work-Energy problem involving pulley

  1. Feb 21, 2012 #1
    1. The problem statement, all variables and given/known data
    The system shown is at rest when a constant 250N force is applied to block A. Neglecting masses of pulleys and friction, determine:
    a) Velocity of block B after block A moved 2 m.
    b) The tension in the cable
    http://imageshack.us/photo/my-images/259/photoivl.jpg/

    2. Relevant equations

    Fnet = ma.

    U = mgh

    KE = 1/2 mv^2

    Ui + KEi + W = Uf + KEf, where subindex i and f mean initial and final states.

    3. The attempt at a solution

    I have completed this problem, but I'm not sure if this is correct. The professor said this particular problem can be done by taking both masses and the ropes as a single system and also by taking each mass as its own, separate system. I did took each as a separate system.

    First, I drew a free body diagram for block A. The net force in the x direction, assuming left is negative gives me:

    F - T = ma_A
    1------------------250N - T = ma_A

    The relationship between the position of block A and block B was obtained and determined to be:
    X_A + 3X_B = constant. Taking first time derivative:
    V_A + 3V_B = 0
    V_A = - 3V_B. Second time derivative:
    2--------------a_A = -3a_B

    Using newton's law for the y direction on block B:
    up positive: 3T - W = m*a_B
    3-----------3T - mg = m*a_B

    Solving 1 for acceleration a_A, then using equation 2 to leave as a_B = form:
    250-T/m_a = -3a_B
    4----- -250+T/(3m_a) = a_B

    Using 4 to solve for tension in 3:
    3T - m_b*g = -m_b*250/(3*m_a) + m_b * T/(3m_a)

    Subsituting for values of mass of b, g, mass of a and solving for T:
    T(3-25/90) = 25*9.81 - 25*250/(3*30)
    T = 64.6N


    Then, with that tension, using the work energy theorem on block A:

    Wtension + Wconst force = 1/2 m_a V_a^2
    250*2 - 129.2 = 1/2 *30va^2
    V_a = 4.97 m /s <------ left.

    Using relationship:
    V_a = -3Vb
    V_b = 1.66 m / s Upwards.





    Does that seem right...? Thanks a ton. :]
     
  2. jcsd
  3. Feb 22, 2012 #2

    ehild

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    Both bodies move along the rope, there is no reason to distinguish between x and y components of velocity. When A moves to the left, B will move vertically upward. Taking "to the left" as positive direction for A, "up" is positive for B. You correctly figured out that the acceleration of A is three times the acceleration of B. So aA=3aB.
    The equations for the accelerations and forces you wrote are correct with this choice of the direction of accelerations:

    25-T=mAaA
    and
    3T-mBg=mBaB

    Proceed from here, but do not forget the parentheses. You had at least one error by ignoring them.

    ehild
     
  4. Feb 23, 2012 #3
    I tried it again and more carefully this time, but used only the magnitudes of acceleration in F =ma equations. I ended up getting a tension of about 96N I think. I can try the calculation here on my phone. XD

    250-T = 25a_a
    3T - 30g = 30 a_b

    a_A = 3a_B

    Plugging 3a_B in first equation:
    250 - T = 75a_B

    1/75 (250 - T) = a_B =1/30(3T - 30g)
    250/75 -T/75 = T/10 - g

    T/75 + T/10 = 250/75 + g
    T + 7.5T = 250 + 75g
    Oh great now I get T = 116N :/
     
  5. Feb 23, 2012 #4
    No wait I think that's what I got for my original calculation actually. And velocity of block A comes out to be 4.63 m/s to the left using the W-E theorem and the speed of block B a third of that value.
     
  6. Feb 23, 2012 #5

    ehild

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    The masses are interchanged in the equations above: ma = 30 kg and mb=25 kg. Your first calculation was correct, T=95.9 N.
     
  7. Feb 23, 2012 #6

    ehild

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    I got va=4.63 m/s, vb=1.51 m/s, but it seems all right otherwise.

    ehild
     
  8. Feb 23, 2012 #7
    Thanks a ton. :]
     
  9. Feb 23, 2012 #8

    ehild

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    You are welcome. It was a nice solution of a nice problem:smile:


    ehild
     
  10. Feb 24, 2012 #9
    thepatient can you explain how you arrived at 3T-W=mbab ?
     
  11. Feb 24, 2012 #10

    ehild

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  12. Feb 24, 2012 #11
  13. Feb 24, 2012 #12

    ehild

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    It is the weight downwards and three times the tension in the string upwards, as the block is held by three threads of the string.
    What do you mean on "the normal of its weight"?

    ehild
     
  14. Feb 24, 2012 #13
    Uh sorry my bad, I was referring to the normal reaction on an object kept on a table lol xD here it doesn't apply sorry
     
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